Question

Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 24 N, and F3 = 19 N. The radial distances are R2 = 0.22 m and R3 = 0.10 m. The forces F2 and F3 are perpendicular to the radial lines labeled R2 and R3. The moment of inertia of the cylinder is 12MR22. Find the magnitude of the angular acceleration of the cylinder about the axis of rotation.

Answer 1

The angular acceleration is 11.66 rad/s²

Step-by-step explanation:

Step 1: Given data

Three forces are applied to a solid cylinder of mass 12 kg

F1 = 15 N

F2 = 24 N

F3 = 19 N

R2 = 0.22m

R3 = 0.10m

Step 2: Find the magnitude of the angular acceleration

I = ½mr² = ½ * 12kg * (0.22m)² = 0.29 kg*m²

torque τ = I*α

τ = F2*R2 - F1*R1 = 24N*0.22m - 19N*0.10m = 3.38 N*m

This means

I = ½mr² = 0.29 kg*m²

τ = I*α = 3.38 N*m

OR

0.29 kg*m² * α = 3.38 N*m

α = 11.655 rad/s² ≈11.66 rad/s²

The angular acceleration is 11.66 rad/s²